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Math Help - Implicit Differentiation #3

  1. #1
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    Implicit Differentiation #3

    Find dy/dx for the following:

    y^3=x^3/(2-xy)

    This is my work.. can someone check to see if I have the correct answer and method? Thanks

    d/dx (3y^2)=(3x^2)(2-xy)-x^3[(-1)y+x(1)dy/dx]/(2-xy)^2

    d/dx (3y^2)=(3x^2)(2-xy)+x^3y-x(dy/dx)]/(2-xy)^2

    (2-xy)^2(3y^2)(dy/dx)+x (dy/dx)=3x^2(2-xy)+x^3y
    dy/dx[(2-xy)^2(3y^2)+x]=3x^2(2-xy)+x^3y
    dy/dx=[3x^2(2-xy)+x^3y]/[(2-xy)^2(3y^2)+x]
    Last edited by mr fantastic; August 12th 2009 at 04:00 PM. Reason: Changed post title
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  2. #2
    Member eXist's Avatar
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    I see a mistake on the RHS on the first step:

    You have \frac{(2-xy)(3x^2) - x^3(-y+xdy/dx)}{(2-xy)^2}

    Check the signs again, I believe it should be: \frac{(2-xy)(3x^2) - x^3(-y-xdy/dx)}{(2-xy)^2}
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  3. #3
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    Quote Originally Posted by eXist View Post
    I see a mistake on the RHS on the first step:

    You have \frac{(2-xy)(3x^2) - x^3(-y+xdy/dx)}{(2-xy)^2}

    Check the signs again, I believe it should be: \frac{(2-xy)(3x^2) - x^3(-y-xdy/dx)}{(2-xy)^2}

    d/dx (3y^2)=(3x^2)(2-xy)-x^3[(-1)y-x(1)dy/dx]/(2-xy)^2

    d/dx (3y^2)=(3x^2)(2-xy)+x^3y+x(dy/dx)]/(2-xy)^2

    (2-xy)^2(3y^2)(dy/dx)-x (dy/dx)=3x^2(2-xy)+x^3y
    dy/dx[(2-xy)^2(3y^2)-x]=3x^2(2-xy)+x^3y
    dy/dx=[3x^2(2-xy)+x^3y]/[(2-xy)^2(3y^2)-x]

    is this correct now?
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  4. #4
    Member eXist's Avatar
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    Look at this part right here for now:
    -x^3(-y-xdy/dx)

    And look at your result:
    x^3y+xdy/dx

    Are you sure?
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  5. #5
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    Quote Originally Posted by eXist View Post
    Look at this part right here for now:
    -x^3(-y-xdy/dx)

    And look at your result:
    x^3y+xdy/dx

    Are you sure?

    so should it be this..
    final result:

    dy/dx=[3x^2(2-xy)+x^3y]/(2-xy)^2(3y^2)-x^4

    ?correct now?
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  6. #6
    Member eXist's Avatar
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    Looks good now ! Just to make sure others understand, make sure to use parenthesis on the bottom:

    Yours:
    dy/dx=[3x^2(2-xy)+x^3y]/(2-xy)^2(3y^2)-x^4

    Yours with parenthesis/brackets:
    dy/dx=[3x^2(2-xy)+x^3y]/[(2-xy)^2(3y^2)-x^4]

    I'm sure that's what you meant, seeing how your work is correct now.
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